Since gravity is the only external force acting on the cylinder that is capable of doing any work, we can ne the equilibrium and stability of the system by considering the potential energy function. We have ZC0-RO sin o, where aco is the value of zc when 0=0. Thus, since d=CG, we have V=mgaG= mg(2c +dsin 0)=mg(aco- RO sin o +dsin 0) The equilibrium points are given by VV=0, but, in this case, since the position of the system is uniquely determined by a single coordinate, e.g. 0, we can write dv/de mg(-Rsin+d cos 8)=0, or, cos 0=(Rsin o)/d. If d< Rsin o, there will be no equilibrium positions. On the other hand, if d> Rsin o, then ge cos[(Rsin o)/d is an equilibrium point. We note that if geq. is an equilibrium point, then -Beg. is also an equilibrium point (i.e. cos 0= cos(-0) In order to study the stability of the equilibrium points, we need to determine whether the potential energy is a maximum or a minimum at these points. Since dv/d82= - sin 0, we have that when geq. <0 hen d-v/de2>0 and the potential energy is a minimum at that point. Consequently, for geq.< 0, the quilibrium is stable. On the other hand, for 6eg.>0, the equilibrium point is unstable ADDITIONAL READING J. L. Meriam and L G. Kraige, Engineering Mechanics, DYNAMICS, 5th EditionSince gravity is the only external force acting on the cylinder that is capable of doing any work, we can examine the equilibrium and stability of the system by considering the potential energy function. We have zC = zC0 − Rθ sin φ, where zC0 is the value of zC when θ = 0. Thus, since d = |CG|, we have, V = mgzG = mg(zC + d sin θ) = mg(zC0 − Rθ sin φ + d sin θ) . The equilibrium points are given by ∇V = 0, but, in this case, since the position of the system is uniquely determined by a single coordinate, e.g. θ, we can write ∇V = dV dθ ∇θ , which implies that, for equilibrium, dV /dθ = mg(−R sin φ + d cos θ) = 0, or, cos θ = (R sin φ)/d. If d < R sin φ, there will be no equilibrium positions. On the other hand, if d ≥ R sin φ, then θ eq. = cos−1 [(R sin φ)/d] is an equilibrium point. We note that if θ eq. is an equilibrium point, then −θ eq. is also an equilibrium point (i.e. cos θ = cos(−θ)). In order to study the stability of the equilibrium points, we need to determine whether the potential energy is a maximum or a minimum at these points. Since d 2V /dθ2 = −mgd sin θ, we have that when θ eq. < 0, then d 2V /dθ2 > 0 and the potential energy is a minimum at that point. Consequently, for θ eq. < 0, the equilibrium is stable. On the other hand, for θ eq. > 0, the equilibrium point is unstable. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 6/6, 6/7 6